© 2010 pearson prentice hall. all rights reserved. chapter 7 algebra: graphs, functions, and linear...

© 2010 Pearson Prentice Hall. All rights reserved.

CHAPTER 7

Algebra: Graphs, Functions, and Linear

Systems

© 2010 Pearson Prentice Hall. All rights reserved. 2

7.3

Systems of Linear Equations in Two Variables


Objectives1. Decide whether an ordered pair is a solution of a

linear system.

2. Solve linear systems by graphing.

3. Solve linear systems by substitution.

4. Solve linear systems by addition.

5. Identify systems that do not have exactly one ordered-pair solution.

6. Solve problems using systems of linear equations.

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Systems of Linear Equations & Their Solutions

• Two linear equations are called a system of linear equations or a linear system.

• A solution to a system of linear equations in two variables is an ordered pair that satisfies both equations in the system.

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Determine whether (1,2) is a solution of the system:

2x – 3y = −4

2x + y = 4

Example 1: Determining Whether an Ordered Pair is a Solution of a System

Solution: Because 1 is the x-coordinate and 2 is the y-coordinate of (1,2), we replace x with 1 and y with 2.

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2x – 3y = − 4 2x + y = 4

2(1) – 3(2) = −4 2(1) + 2 = 4

2 – 6 = − 4 2 + 2 = 4

− 4 = − 4, TRUE 4 = 4, TRUE

The pair (1,2) satisfies both equations; it makes each equation true. Thus, the pair is a solution of the system.

Example 1 continued

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Graphing 2 lines

• 2x – 3y = -42x + 4 = 3yy = (2/3)x + (4/3)

• 2x + y = 4y = -2x + 4

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Solving Linear Systems by Graphing

• For a system with one solution, the coordinates of the point of intersection of the lines is the system’s solution.

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Example 2: Solving Linear Systems by Graphing

Solve by graphing:

x + 2y = 2

x – 2y = 6.

Solution:

We find the solution by graphing both x + 2y = 2 and x – 2y = 6 in the same rectangular coordinate system. We will use intercepts to graph each equation.

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x-intercept: Set y = 0.

x + 2 · 0 = 2

x = 2

The line passes

through (2,0).

Example 2 Continued

y-intercept: Set x = 0.

0 + 2y = 2

2y = 2

y = 1

The line passes

through (0,1).

We will graph x + 2y = 2 as a blue line.

x + 2y = 2:

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x-intercept: Set y = 0.

x – 2 · 0 = 6

x = 6

The line passes through (6,0).

Example 2 Continued

y-intercept: Set x = 0.

0 – 2y = 6

−2y = 6

y = −3

The line passes through (0,−3).

We will graph x – 2y = 6 as a red line.

x – 2y = 6:

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We see the two graphs

intersect at (4,−1). Hence,

this is the solution to the

system.

We can check this

by substituting in (4,−1) into

each equation and verifying

That the solution is true for both

equations.

We leave this to the student.

Example 2 Continued

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Solving Linear Systems by the Substitution Method

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Solve by the substitution method:

y = −x – 1

4x – 3y = 24.

Solution:

Step 1 Solve either of the equations for one variable in terms of the other. This step has been done for us. The first equation, y = −x – 1, is solved for y in terms of x.

Example 3: Solving a System by Substitution

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Step 2 Substitute the expression from step 1 into the other equation.

This gives us an equation in one variable, namely

4x – 3(−x – 1) = 24.

The variable y has been eliminated.

Example 3 continued

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Step 3 Solve the resulting equation containing one variable. 4x – 3(-x – 1) = 24

4x + 3x + 3 = 24

7x + 3 = 24

7x = 21

x = 3

Step 4 Back-substitute the obtained value into the equation from step1. Since we found x = 3 in step 3, then we back-substitute the x-value into the equation from step 1 to find the y-coordinate.

Example 3 continued

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Step 4 (cont.)

With x = 3 and y = −4, the proposed solution is (3,−4).

Step 5 Check. Use this ordered pair to verify that this solution makes each equation true. We leave this to the student.

Example 3 continued

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Your Turn

• Solve the following system of equations by the substitution method.

• y = 2x + 72x – y = -5

• -4x + y = -112x – 3y = 3

1


Solving Linear Systems by the Addition Method

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Solve by the addition method:

3x + 2y = 48

9x – 8y = −24.

Example 5: Solving a System by the Addition Method

Solution:

Step 1 Rewrite both equations in the form Ax + By = C. Both equations are already in this form. Variable terms appear on the left and constants appear on the right.

Step 2 If necessary, multiply either equation or both equations by appropriate numbers so that the sum of the x-coefficients or the sum of the y-coefficients is 0.

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Example 5 continued

3x + 2y = 48 − 9x – 6y = − 144

9x – 8y = −24 9x – 8y = − 24

Step 3 Add the equations. −14y = −168

Multiply by −3

No Change

Step 4 Solve the equation in one variable. We solve −14y = −168 by dividing both sides by −14.

14 168

14 14 12

y

y

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Step 5 Back-substitute and find the value for the other variable.

3x + 2y = 48

3x + 2(12) = 48

3x + 24 = 48

3x = 24

x = 8

Example 5 continued

Step 6 Check. The solution to the system is (8,12). We can check this by verifying that the solution is true for both equations. We leave this to the student.

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Linear Systems Having No Solution or Infinitely Many Solutions

The number of solutions to a system of two linear equations in two variables is given by one of the following:

Number of Solutions What This Means Graphically

Exactly one ordered-pair solution The two lines intersect at one point.

No Solution The two lines are parallel.

Infinitely many solutions The two lines are identical.

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Solve the system:

4x + 6y = 12

6x + 9y = 12.

Solution: Because no variable is isolated, we will use the addition method.

4x + 6y = 12

6x + 9y = 12

The false statement 0 = 12 indicates that the system has no solution. The solution is the empty set, Ø.

Example 7: A System with no Solution

Multiply by 3

Multiply by -2

Add:

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Solve the system:

y = 3x – 2

15x – 5y = 10.

Solution: Because the variable y is isolated in y = 3x – 2,

the first equation, we will use the substitution method.

Example 8: A System with Infinitely Many Solutions

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The statement 10 = 10 is true. Hence, this indicates that the system has infinitely many solutions.

Example 8 continued

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Modeling with Systems of Equations: Making Money (and Losing It)

Revenue and Cost Functions

A company produces and sells x units of a product.

• Revenue Function R(x) = (price per unit sold)x

• Cost Function C(x) = fixed cost + (cost per unit produced)x

The point of intersection of the graphs of the revenue and cost functions is called the break-even point.

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A company is planning to manufacture radically different wheelchairs. Fixed cost will be $500,000 and it will cost $400 to produce each wheelchair. Each wheelchair will be sold for $600.

a. Write the cost function, C, of producing x wheelchairs.

b. Write the revenue function, R, from the sale of x wheelchairs.

c. Determine the break-even point. Describe what this means.

Example 9: Finding a Break-Even Point

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Solution:

a. The cost function is the sum of the fixed cost and the variable cost.

b. The revenue function is the money generated from the sale of x wheelchairs.

Example 9 Continued

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c. The break even point occurs where the graphs of C and R intersect. Thus, we find this point by solving the system

C(x) = 500,000 + 400x

R(x) = 600x,

Or

y = 500,000 + 400x

y = 600x.

Example 9 continued

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Using substitution, we substitute 600x in for y in the first equation:

600x = 500,000 + 400x

200x = 500,000

x = 2500

Back-substituting 2500 for x in either of the system’s equations (or functions), we obtain

Example 9 continued

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The break-even point is (2500, 1,500,000). This means that the company will break even if it produces and sells 2500 wheelchairs for $1,500,000.

Example 9 continued

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The Profit Function

• The profit, P(x), generated after producing and selling x units of a product is given by the profit function

P(x) = R(x) – C(x),

where R and C are the revenue and cost, respectively.

The profit function, P(x), for the

previous example is

P(x) = R(x) – C(x)

= 600x – (500,000 + 400x)

= 200x – 500,000.

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© 2010 pearson prentice hall. all rights reserved. chapter 7 algebra: graphs, functions, and linear...

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